 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/36c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The one-particle partition function is

z_{1} = \frac{A}{h ^{3}} \int_{R^{3}} e^{- β p^{2} / (2 m)} d^{3} p \int_{0}^{\infty} e^{- β m g z} d z = \frac{A}{λ ^{3} β m g},

where

λ = (2 π ℏ^{2} β / m)^{1/2}

. Thus

Z_{N} = \frac{z _{1}^{N}}{N !}, lo g z_{1} = constant - \frac{5}{2} lo g β .

Canonical differentiation now gives

⟨ E ⟩ = - \partial_{β} lo g Z_{N} = \frac{5 N}{2 β} = \frac{5}{2} N k_{B} T

and

(Δ E)^{2} = \partial_{β}^{2} lo g Z_{N} = \frac{5 N}{2 β ^{2}} .

Therefore

\frac{Δ E}{⟨ E ⟩} = \frac{2}{5 N} .

The factor

5/2

consists of

3/2

1

from gravitational potential energy.

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